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denpristay [2]
3 years ago
6

9(9k + 12n + 6), for k = 7 and n = 5

Mathematics
1 answer:
joja [24]3 years ago
5 0
Using PEMDAS

9(9(7) + 12(5) + 6)
9(63 + 60 + 6) 
9(129)
<span>0

Just kidding, the answer is: 1161
</span>
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Plz help.. I asked this question on English, but no one answered.
enyata [817]

Answer:

Step-by-step explanation:

A. abject - 2. wretched

B. congenial - 6. friendly

C. deliberately - 7. thoughtfully

D. impetuous - 3. impulsive

E. misgiving - 4. doubt

F. perturbation - 5. disturbance

G. pervade - 8. permeate

H. transgress - 1. err

Hope this helps :)

4 0
4 years ago
Last year, a comprehensive report stated that 28% of businesses in the northeast of Ohio were considered highly profitable. This
Alik [6]

Answer:

z=\frac{0.38 -0.28}{\sqrt{\frac{0.28(1-0.28)}{50}}}=1.575  

p_v =2*P(z>1.575)=0.115  

So the p value obtained was a very high value and using the significance level given \alpha=0.01 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of businesses were highly profitable is not significantly different from 0.28 or 28%.

Step-by-step explanation:

Data given and notation

n=50 represent the random sample taken

X=19 represent the businesses were highly profitable

\hat p=\frac{19}{50}=0.38 estimated proportion of businesses were highly profitable

p_o=0.28 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion of businesses were highly profitable is different from 0.28 or no, the system of hypothesis is.:  

Null hypothesis:p=0.28  

Alternative hypothesis:p \neq 0.28  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info required we can replace in formula (1) like this:  

z=\frac{0.38 -0.28}{\sqrt{\frac{0.28(1-0.28)}{50}}}=1.575  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.01. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =2*P(z>1.575)=0.115  

So the p value obtained was a very high value and using the significance level given \alpha=0.01 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of businesses were highly profitable is not significantly different from 0.28 or 28%.

4 0
4 years ago
Evaluate the expression. StartFraction 9 factorial Over 3 factorial EndFraction 3 6 60,480 362,874
REY [17]

Answer:

60,480 is the correct answer.

Step-by-step explanation:

First of all, let us have a look at the <em>formula of factorial of a number 'n'</em>:

n! = n \times (n-1) \times (n-2) \times ...... \times 1

i.e. multiply n with (n-1) then by (n-2) upto 1.

<em>Keep on subtracting 1 from the number and keep on multiplying until we reach to 1.</em>

<em></em>

So, 9! can be written as: 9 \times 8 \times 7 \times ...... \times 1

Similarly 3! can be written as: 3 \times 2 \times 1

Re-writing 9 ! :

9 \times 8 \times 7 \times ...... 3 \times 2 \times 1\\\Rightarrow 9 \times 8 \times 7 \times ...... 3 !

Now, the expression to be evaluated:

\dfrac{9!}{3!} = \dfrac{9 \times 8 \times 7 \times ..... \times 3!}{3!}\\\Rightarrow 9 \times 8 \times 7 \times 6 \times 5 \times 4\\\Rightarrow 60480

5 0
3 years ago
Read 2 more answers
jon has $442 in his bank account. he spend $29 on gas and $68 on dinner. how much money is left in jon's bank account after thes
skad [1K]

Answer:

$345

Step-by-step explanation:

Take $442 and subtract it to $29 and $68 to get $345.

8 0
3 years ago
I need the help, you need the points
Alex_Xolod [135]
(2.75 g -2.699 g)/(2.699 g) * 100% = 1.89%

_____
Perhaps you can tell that the general formula is
.. (measured value - actual value)/(actual value) * 100% = % error
3 0
3 years ago
Read 2 more answers
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